Question: Given that $x$ is real and $x^3+\frac{1}{x^3}=52$, find $x+\frac{1}{x}$.
Solution: We know that $$\left(x+\frac{1}{x}\right)^3=x^3+3(x^2)\left(\frac{1}{x}\right)+3(x)\left(\frac{1}{x}\right)^2+\left(\frac{1}{x}\right)^3=x^3+\frac{1}{x^3}+3\left(x+\frac{1}{x}\right).$$Let $x+\frac{1}{x}=a$. Then our equation is $a^3=x^3+\frac{1}{x^3}+3a$. We know $x^3+\frac{1}{x^3}=52$, so we have $a^3=52+3a$ or $a^3-3a-52=0$. By the rational root theorem, the possible roots of this polynomial equation are the divisors of 52 as well as their negatives: $\pm1, \pm 2, \pm 4, \pm 13, \pm 26, \pm 52$.  Both $\pm1$ and $\pm2$ are easy to check by substitution. For $\pm 4$ we can use synthetic division (or substitution), and we find that that $a=4$ is a root. (We could also see this by inspection by writing $a^3-3a=52$ and noting that $4$ works.)

Are there other solutions? Use synthetic division to divide:

\begin{tabular}{c|cccc}
$4$&$1$&$0$&$-3$&$-52$\\
$$&$\downarrow$&$4$&$16$&$52$\\ \hline
$$&$1$&$4$&13$$&$0$
\end{tabular}
The quotient is $a^2+4a+13$, so $a^3-3a-52 = (a-4)(a^2+4a+13)$. The discriminant of $a^2+4a+13$ is $4^2-4(1)(13)=16-52=-36$, which is negative, so there are no other real solutions for $a$. If $x$ is real, $a$ must be real, so we conclude that there are no other values of $x+\frac{1}{x}$. Thus $x+\frac{1}{x}=a=\boxed{4}$.